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Q.
A solution of $10 \,ml \,\frac{M}{10} FeSO_4$ was titrated with $KMnO_4$ solution in acidic medium. The amount of $KMnO_4$ used will be :
Redox Reactions
Solution:
Given: The $10\, ml$ of $\frac{ M }{10} FeSO _{4}$ is titrated with $KMnO _{4} .$
To Find: Amount of $KMnO _{4}$ used.
Reaction:
$MnO _{4}^{-}+5 Fe ^{2+}+8 H ^{+} \rightarrow 5 Fe ^{3+}+ Mn ^{2+}+4 H _{2} O$
Hence, 1 mole of $KMnO _{4}$ is required to titrate 5 moles of $FeSO _{4}$.
Moles of $FeSO _{4}$ titrated =Molality $\times$ Volume
$=10\, ml \times \frac{M}{10}=1\, m$ mole.
The number of moles of $KMnO _{4}$
used $=\left(\frac{\text { Moles of } FeSO _{4}}{5}\right)$
$=\frac{1 mill \text { mole }}{5}=0.2\, m$ mole.
So, $0.2\, m$ mole $=10\, ml \times 0.02\, M\, KMnO _{4}$.
Therefore, $10\, ml$ of $0.02\, M\, KMO _{4}$ is required to titrate the $FeSO _{4}$ solution.